Consider the decomposition of barium carbonate: BaCO3(s) -> BaO(s) + CO2(g). Usi
ID: 721965 • Letter: C
Question
Consider the decomposition of barium carbonate: BaCO3(s) -> BaO(s) + CO2(g). Using data from the thermodynamic quantities at 298.15K table, calculate the equilibrium pressure of CO2 at (a) 298K and (b) 1100K.I have the answers, but need to understand the concept and the steps. Thanks!
The answers provided are: deltaH^degree = 269.3 KJ, deltaS^degree = 0.1719 J/K, (a)PCO2 = 6 x 10^-39 atm, and (b) PCO2 = 1.6 x 10^-4 atm
Explanation / Answer
By definitions dHrxn = dHproducts - dHreactants dHrxn you found = 269.3 kJ dSrxn is found in the same way (products - reactants) We basically set up our products and reactants: BaCO3(s) -> BaO(s) + CO2(g) dAnything = (BaO + CO2) - (BaCO3) The thermodynamic quantities in the table you have merely plugged in those values into the above equations to find the dS and dH respectively. For instance: dH = (Hformation of BaO + Hformation of CO2) - (Hformation of BaCO3) For partial pressures use the equation G = -RTlnK G = H- TS Plug in your values, for example in part A you would use, G = 269.3 - 298(0.1719) Once you have G you divide by -RT Where lnK = -G/RT Once you solve -G/RT then you set that side to the power of e to get rid of the logarithm Where K = e^(-G/RT) That'll get you K, now this is a subtle note for solids, in this case BaCO3 and BaO, assume their partial pressures are 1. K = Products / Reactants = pCO2 * pBaO / BaCO3 So essentially K is your pCO2